ex: Solved some other exercises, added code snippets, and more
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{
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"My exercise environment": {
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"scope": "latex,tex",
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"prefix": "myexercise",
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"body": [
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"\\begin{myexercise}",
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"\t$1",
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"\\end{myexercise}"
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],
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},
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"My answer environment": {
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"scope": "latex,tex",
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"prefix": "myanswer",
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"body": [
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"\\begin{myanswer}",
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"\t$1",
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"\\end{myanswer}"
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],
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},
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}
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@ -2,5 +2,5 @@
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JUNK="*.aux *.bbl *.bcf *.blg *.log *.out *.run.xml *.toc *.lof *.lot *.fls *.fdb_latexmk *.synctex* *.ptc *.ind *.ilg *.idx *.glo *.glsdefs *.glg *.gls"
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JUNK="*.aux *.bbl *.bcf *.blg *.log *.out *.run.xml *.toc *.lof *.lot *.fls *.fdb_latexmk *.synctex* *.ptc *.ind *.ilg *.idx *.glo *.glsdefs *.glg *.gls"
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for item in $JUNK; do
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for item in $JUNK; do
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find . -name "$item" -exec rm {} \;
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find . -name "$item" -exec rm {} \;
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done
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done
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%----------------------------------------
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%----------------------------------------
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\newcommand{\includechapters}[2]{%
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\newcommand{\includechapters}[2]{%
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\foreach \i in {0, ..., #2} {%
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\foreach \i in {0, ..., #2} {%
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\edef\FileName{content/chapters/#1/\i}% The % here are necessary to eliminate any
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\edef\FileName{content/chapters/#1/\i}% The % here are necessary to eliminate any
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\IfFileExists{\FileName}{% spurious spaces that may get inserted
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\IfFileExists{\FileName}{% spurious spaces that may get inserted
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\input{\FileName}% at these points
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\input{\FileName}% at these points
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}
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}
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}
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}
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}
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}
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@ -8,60 +8,58 @@
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\begin{myexercise}{4}{18}
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\begin{myexercise}{4}{18}
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Let $F_n$ be the $n$-th Fibonacci number, defined by the recursion relation
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Let $F_n$ be the $n$-th Fibonacci number, defined by the recursion relation
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\[
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\[
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\begin{cases}
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\begin{cases}
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F_0 = 1, \\
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F_0 = 1, \\
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F_1 = 1, \\
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F_1 = 1, \\
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F_n = F_{n-1} + F_{n-2}, \quad n \geq 2.
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F_n = F_{n-1} + F_{n-2}, \quad n \geq 2.
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\end{cases}
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\end{cases}
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\]
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\]
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Let $\phi = \frac{1+\sqrt{5}}{2}$ be the golden ratio.
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Let $\phi = \frac{1+\sqrt{5}}{2}$ be the golden ratio.
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We have
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We have
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\begin{equation}
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\begin{equation}
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F_n \leq \phi^{n-1}
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F_n \leq \phi^{n-1}
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\label{eq:fibo_leq_golden_ratio}
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\label{eq:fibo_leq_golden_ratio}
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\end{equation}
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\end{equation}
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$\forall n \in \mathbb{N}$.
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$\forall n \in \mathbb{N}$.
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Prove that, in addition to Eq. \ref{eq:fibo_leq_golden_ratio}, $F_n \geq \phi^{n-2}$.
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Prove that, in addition to Eq. \ref{eq:fibo_leq_golden_ratio}, $F_n \geq \phi^{n-2}$.
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\begin{myanswer}
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\begin{myanswer}
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Let $P(n)$ be the assertion that $F_n \geq \phi^{n-2}$.
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Let $P(n)$ be the assertion that $F_n \geq \phi^{n-2}$.
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We proceed by recursion on $n$.
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We proceed by recursion on $n$.
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\begin{equation}
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\begin{equation}
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1 + \phi = \phi^2
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1 + \phi = \phi^2
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\label{eq:golden_ratio_squared}
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\label{eq:golden_ratio_squared}
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\end{equation}
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\end{equation}
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\emph{Initiation} For $n = 0$: $\phi^{0-2} = \phi^{-2} = \frac{1}{\phi^2}$. As $\phi^2 > 1$ . $\frac{1}{\phi^2} < 1$, so $P(0)$ is true.
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\emph{Initiation} For $n = 0$: $\phi^{0-2} = \phi^{-2} = \frac{1}{\phi^2}$. As $\phi^2 > 1$ . $\frac{1}{\phi^2} < 1$, so $P(0)$ is true.
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For $n = 1$:
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For $n = 1$:
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$\phi^{1-2} = \phi^{-1} = \frac{1}{\phi}$. As $\phi > 1$, $\frac{1}{\phi} < 1$, so $P(1)$ is true.
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$\phi^{1-2} = \phi^{-1} = \frac{1}{\phi}$. As $\phi > 1$, $\frac{1}{\phi} < 1$, so $P(1)$ is true.
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\emph{Heredity}
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\emph{Heredity}
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Let $P(n)$ and $P(n-1)$ be true. We have to prove $P(n+1)$.
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Let $P(n)$ and $P(n-1)$ be true. We have to prove $P(n+1)$.
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\begin{align*}
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\begin{align*}
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P(n) \wedge P(n-1) & \Leftrightarrow F_n \geq \phi^{n-2} \wedge F_{n-1} \geq \phi^{n-1-2}\\
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P(n) \wedge P(n-1) & \Leftrightarrow F_n \geq \phi^{n-2} \wedge F_{n-1} \geq \phi^{n-1-2}\\
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& \Leftrightarrow F_{n} + F_{n-1} \geq \phi^{n-2} + \phi^{n-3} \\
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& \Leftrightarrow F_{n} + F_{n-1} \geq \phi^{n-2} + \phi^{n-3} \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} (1 + \phi) \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} (1 + \phi) \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} \phi^2 \qquad \text{using Eq. \ref{eq:golden_ratio_squared}} \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-3} \phi^2 \qquad \text{using Eq. \ref{eq:golden_ratio_squared}} \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-1} \\
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& \Leftrightarrow F_n + F_{n-1} \geq \phi^{n-1} \\
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& \Leftrightarrow F_{n+1} \geq \phi^{n+1-2} \\
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& \Leftrightarrow F_{n+1} \geq \phi^{n+1-2} \\
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& \Leftrightarrow P(n+1)
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& \Leftrightarrow P(n+1)
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\end{align*}
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\end{align*}
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\emph{Conclusion} $P(0)$ and $P(1)$ are true. $P(n) \wedge P(n-1)$ true implies $P(n+1)$ true. So $P(n)$ is true for all $n \in \mathbb{N}$.
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\emph{Conclusion} $P(0)$ and $P(1)$ are true. $P(n) \wedge P(n-1)$ true implies $P(n+1)$ true. So $P(n)$ is true for all $n \in \mathbb{N}$.
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\end{myanswer}
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\end{myanswer}
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\end{myexercise}
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\end{myexercise}
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\begin{myexercise}{5}{19}
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\begin{myexercise}{5}{19}
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A prime number is an integer $> 1$ that has no exact divisors other than $1$ and itself.
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A prime number is an integer $> 1$ that has no exact divisors other than $1$ and itself.
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@ -71,17 +69,17 @@ Using this definition and mathematical induction, prove that every integer $> 1$
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Let $P(n)$ be the assertion that $n$ can be written as a product of prime numbers or is a prime.
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Let $P(n)$ be the assertion that $n$ can be written as a product of prime numbers or is a prime.
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\begin{proof}
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\begin{proof}
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Let $N > 1$ be the smallest integer such that $P(N)$ is false.
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Let $N > 1$ be the smallest integer such that $P(N)$ is false.
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As $\forall N' < N$, $N' > 1$, $P(N')$ is true.
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As $\forall N' < N$, $N' > 1$, $P(N')$ is true.
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We have either of the following assertions:
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We have either of the following assertions:
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\begin{itemize}
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\begin{itemize}
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\item $N$ is a prime number;
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\item $N$ is a prime number;
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\item there exists $m$ and $n$ below $N$ such that, $m \times n = N$.
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\item there exists $m$ and $n$ below $N$ such that, $m \times n = N$.
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As $m$ and $n$ are below $N$, they satisfies $P$, so they are either primes or a product of primes. So $mn = N$ is also a product of primes.
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As $m$ and $n$ are below $N$, they satisfies $P$, so they are either primes or a product of primes. So $mn = N$ is also a product of primes.
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\end{itemize}
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\end{itemize}
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We have a contradiction, and $P(N)$ must be true.
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We have a contradiction, and $P(N)$ must be true.
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\end{proof}
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\end{proof}
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\end{myanswer}
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\end{myanswer}
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@ -115,156 +113,297 @@ T_n = \frac{1}{2} n (n + 1)
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\end{equation}
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\end{equation}
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\begin{proof}
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\begin{proof}
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Let us prove the formula.
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Let us prove the formula.
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Let $P(n)$ be the proposition that Eq. \ref{eq:ex7_half_factor_n_n+1} is correct for $T_n$.
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Let $P(n)$ be the proposition that Eq. \ref{eq:ex7_half_factor_n_n+1} is correct for $T_n$.
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By recursion on $n$
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By recursion on $n$
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\emph{Initiation}
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\emph{Initiation}
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For $n = 1$, $1^2 = 1$ and $\frac{1}{2} 1 \times (1+1) = 1$, so $P(n)$ is true.
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For $n = 1$, $1^2 = 1$ and $\frac{1}{2} 1 \times (1+1) = 1$, so $P(n)$ is true.
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\emph{Heredity}
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\emph{Heredity}
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Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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\begin{align*}
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\begin{align*}
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P(n) & \Leftrightarrow T_n = \frac{1}{2} n(n+1) \\
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P(n) & \Leftrightarrow T_n = \frac{1}{2} n(n+1) \\
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& \Leftrightarrow \sum_{i=0}^n (-1)^i (n-i)^2 = \frac{1}{2} n(n+1)
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& \Leftrightarrow \sum_{i=0}^n (-1)^i (n-i)^2 = \frac{1}{2} n(n+1)
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\end{align*}
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\end{align*}
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\unfinished
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\unfinished
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\end{proof}
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\end{proof}
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\end{myanswer}
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\end{myanswer}
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\end{myexercise}
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\end{myexercise}
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\begin{myexercise}{8}{19}
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\begin{myexercise}{8}{19}
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(a) Prove the following theorem of Nicomachus by induction:
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(a) Prove the following theorem of Nicomachus by induction:
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$1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$, $4^3 = 13 + 15 + 17 + 19$, etc.
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$1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$, $4^3 = 13 + 15 + 17 + 19$, etc.
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(b) Use this result to prove the remarkable formula $1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$
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(b) Use this result to prove the remarkable formula $1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$
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\begin{myanswer}
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\begin{myanswer}
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(a) The theorem states the following:
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(a) The theorem states the following:
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For all $n \in \mathbb{N}$ we have
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For all $n \in \mathbb{N}$ we have
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\[
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\[
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n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|
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n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|
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\]
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\]
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\begin{proof}
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\begin{proof}
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We proceed by recursion on $n$.
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We proceed by recursion on $n$.
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Let $P(n)$ be the proposition ``$n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|$''.
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Let $P(n)$ be the proposition ``$n^3 = \sum_{i = 1}^n | n (n - 1) - 1 + 2i|$''.
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\emph{Initiation} For $n = 1$, $1^3 = 1^2$, so $P(1)$ is true.
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\emph{Initiation} For $n = 1$, $1^3 = 1^2$, so $P(1)$ is true.
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\emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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\emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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\begin{align*}
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\begin{align*}
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P(n) \Leftrightarrow n^3 &= \sum_{i = 1}^n n (n - 1) - 1 + 2i \\
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P(n) \Leftrightarrow n^3 &= \sum_{i = 1}^n n (n - 1) - 1 + 2i \\
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% &\Leftrightarrow (n+1)^3 &= \sum_{i = 1}^n n^2 - n - 1 + 2i \\
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% &\Leftrightarrow (n+1)^3 &= \sum_{i = 1}^n n^2 - n - 1 + 2i \\
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\end{align*}
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\end{align*}
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\end{proof}
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\end{proof}
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\unfinished
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\unfinished
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(b)
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(b)
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\begin{theorem}[Formula of Nicomachus]
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\begin{theorem}[Formula of Nicomachus]
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The sum of the cubes of the first $n$ natural numbers is equal to the square of the sum of the first $n$ natural numbers.
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The sum of the cubes of the first $n$ natural numbers is equal to the square of the sum of the first $n$ natural numbers.
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\end{theorem}
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\end{theorem}
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\begin{equation}
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\begin{equation}
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\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2
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\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2
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\end{equation}
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\end{equation}
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\begin{proof}
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\begin{proof}
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Let $P(n)$ be the proposition ``$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$''.
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Let $P(n)$ be the proposition ``$\sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2$''.
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$P(n) \Longleftrightarrow \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
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$P(n) \Longleftrightarrow \sum_{k=1}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2$
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because $\sum_{i=1}^n i = \frac{n(n+1)}{2}$.
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because $\sum_{i=1}^n i = \frac{n(n+1)}{2}$.
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We want to prove that: $\sum_{k=1}^{n+1} k^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$
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We want to prove that: $\sum_{k=1}^{n+1} k^3 = \left(\frac{(n+1)(n+2)}{2}\right)^2$
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\begin{align*}
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\begin{align*}
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\sum_{k=1}^{n+1} k^3 &= \frac{n^2 (n+1)^2}{2^2} + (n+1)^3 \\
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\sum_{k=1}^{n+1} k^3 &= \frac{n^2 (n+1)^2}{2^2} + (n+1)^3 \\
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&= (n+1)^2 \left(\frac{n^2}{4} + n + 1\right) \\
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&= (n+1)^2 \left(\frac{n^2}{4} + n + 1\right) \\
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&= (n+1)^2 \cdot \frac{n^2 + 4n + 4}{4} \\
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&= (n+1)^2 \cdot \frac{n^2 + 4n + 4}{4} \\
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&= \frac{(n+1)^2(n+2)^2}{2^2} \\
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&= \frac{(n+1)^2(n+2)^2}{2^2} \\
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& \Updownarrow \\
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& \Updownarrow \\
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& P(n+1)
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& P(n+1)
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\end{align*}
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\end{align*}
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\emph{Conclusion}
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\emph{Conclusion}
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$P(1)$ is true because $1^3 = 1$ and $\left(\frac{1(1+1)}{2}\right)^2 = 1$.
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$P(1)$ is true because $1^3 = 1$ and $\left(\frac{1(1+1)}{2}\right)^2 = 1$.
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$P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
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$P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
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\end{proof}
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\end{proof}
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\end{myanswer}
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\end{myanswer}
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\end{myexercise}
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\end{myexercise}
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\begin{myexercise}{9}{19}
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\begin{myexercise}{9}{19}
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Prove by induction that if $0 < a < 1$ then $(1 - a)^n \geq 1 - na$.
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Prove by induction that if $0 < a < 1$ then $(1 - a)^n \geq 1 - na$.
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\begin{myanswer}
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\begin{myanswer}
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We proceed by recursion on $n$.
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We proceed by recursion on $n$.
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Let $P(n)$ be the proposition ``$(1 - a)^n \geq 1 - na$''.
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Let $P(n)$ be the proposition ``$(1 - a)^n \geq 1 - na$''.
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\emph{Initiation} For $n = 0$,
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\emph{Initiation} For $n = 0$,
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$(1 - a)^n = 1 = 1 - na$, so $P(0)$ is true.
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$(1 - a)^n = 1 = 1 - na$, so $P(0)$ is true.
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\emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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\emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
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We want to prove that $(1 - a)^{n+1} \geq 1 - (n+1)a$.
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We want to prove that $(1 - a)^{n+1} \geq 1 - (n+1)a$.
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\begin{align*}
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\begin{align*}
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P(n) \Leftrightarrow (1-a)^n &\geq 1-na \\
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P(n) \Leftrightarrow (1-a)^n &\geq 1-na \\
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(1-a)^n (1-a) & \geq (1 - na) (1-a) \qquad \text{$1-a > 0 $} \\
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(1-a)^n (1-a) & \geq (1 - na) (1-a) \qquad \text{$1-a > 0 $} \\
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||||||
(1-a)^{n+1} & \geq 1 - na - a + na \\
|
(1-a)^{n+1} & \geq 1 - na - a + na \\
|
||||||
& \geq 1 - (n+1)a - a \\
|
& \geq 1 - (n+1)a - a \\
|
||||||
& \geq 1 - (n+1)a \qquad \text{because $a > 0$}
|
& \geq 1 - (n+1)a \qquad \text{because $a > 0$}
|
||||||
& \Rightarrow P(n+1)
|
& \Rightarrow P(n+1)
|
||||||
\end{align*}
|
\end{align*}
|
||||||
|
|
||||||
\emph{Conclusion}
|
\emph{Conclusion}
|
||||||
$P(1)$ is true.
|
$P(1)$ is true.
|
||||||
|
|
||||||
$P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
|
$P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all $n \in \mathbb{N}_+$.
|
||||||
\end{myanswer}
|
\end{myanswer}
|
||||||
\end{myexercise}
|
\end{myexercise}
|
||||||
|
|
||||||
% \begin{myexercise}{10}{19}
|
% \begin{myexercise}{10}{19}
|
||||||
% Prove by induction that if $n \geq 10$ then $2^n > n^3$.
|
% Prove by induction that if $n \geq 10$ then $2^n > n^3$.
|
||||||
|
|
||||||
% \begin{myanswer}
|
% \begin{myanswer}
|
||||||
% We proceed by recursion on $n$.
|
% We proceed by recursion on $n$.
|
||||||
|
|
||||||
% Let $P(n)$ be the proposition ``$2^n > n^3$''.
|
% Let $P(n)$ be the proposition ``$2^n > n^3$''.
|
||||||
|
|
||||||
% \emph{Initiation} For $n = 10$,
|
% \emph{Initiation} For $n = 10$,
|
||||||
|
|
||||||
% $2^{10} = 1024 > 1000 = 10^3$, so $P(10)$ is true.
|
% $2^{10} = 1024 > 1000 = 10^3$, so $P(10)$ is true.
|
||||||
|
|
||||||
% \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
|
% \emph{Heredity} Suppose $P(n)$ true, let us prove that $P(n+1)$ is also true.
|
||||||
|
|
||||||
% We want to prove that $2^{n+1} > (n+1)^3$.
|
% We want to prove that $2^{n+1} > (n+1)^3$.
|
||||||
|
|
||||||
% \begin{align*}
|
% \begin{align*}
|
||||||
% P(n) \Leftrightarrow 2^n &> n^3 \\
|
% P(n) \Leftrightarrow 2^n &> n^3 \\
|
||||||
% 2^n \times 2 &> n^3 \times 2 \qquad \text{because $2 > 0$} \\
|
% 2^n \times 2 &> n^3 \times 2 \qquad \text{because $2 > 0$} \\
|
||||||
% 2^{n+1} &> n^3 + n^3 \\
|
% 2^{n+1} &> n^3 + n^3 \\
|
||||||
% \end{align*}
|
% \end{align*}
|
||||||
|
|
||||||
% $n^3 = n^2 \times n$
|
% $n^3 = n^2 \times n$
|
||||||
|
|
||||||
% \emph{Conclusion}
|
% \emph{Conclusion}
|
||||||
% $P(10)$ is true.
|
% $P(10)$ is true.
|
||||||
|
|
||||||
% $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all integer $n \geq 10$.
|
% $P(n)$ true implies $P(n+1)$ true, so $P(n)$ is true for all integer $n \geq 10$.
|
||||||
% \end{myanswer}
|
% \end{myanswer}
|
||||||
% \end{myexercise}
|
% \end{myexercise}
|
||||||
% Dirty false proof... To be continued
|
% Dirty false proof... To be continued
|
||||||
|
|
||||||
|
\subsection{Numbers, Powers and Logarithms}
|
||||||
|
|
||||||
|
\begin{myexercise}{1}{25}
|
||||||
|
What is the smallest positive rational number?
|
||||||
|
|
||||||
|
\begin{myanswer}
|
||||||
|
The smallest positive rational number is 0.
|
||||||
|
\end{myanswer}
|
||||||
|
\end{myexercise}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{myexercise}{2}{25}
|
||||||
|
Is $1 + 0.23999999999...$ a decimal expansion?
|
||||||
|
|
||||||
|
\begin{myanswer}
|
||||||
|
0.23999999999... is not a decimal number as it has an infinite decimal expansion.
|
||||||
|
|
||||||
|
The decimal expansion of $1 + 0.23999999999...$ is in the form
|
||||||
|
\[
|
||||||
|
1 \cdot 10^0 + 2 \cdot 10^{-1} + 3 \cdot 10^{-2} + 9 \cdot 10^{-3} + \sum_{i=1}^{\infty} 9 \cdot 10^{-3-i}
|
||||||
|
\]
|
||||||
|
|
||||||
|
\textit{Nota bene}: This surely does not answer the question...
|
||||||
|
\end{myanswer}
|
||||||
|
\end{myexercise}
|
||||||
|
|
||||||
|
|
||||||
|
\begin{myexercise}{3}{25}
|
||||||
|
What is $(-3)^{-3}$?
|
||||||
|
|
||||||
|
\begin{myanswer}
|
||||||
|
\[
|
||||||
|
(-3)^{-3} = \frac{1}{(-3)^3} = \frac{1}{(-3)(-3)(-3)} = -\frac{1}{27}
|
||||||
|
\]
|
||||||
|
\end{myanswer}
|
||||||
|
|
||||||
|
\end{myexercise}
|
||||||
|
|
||||||
|
\begin{myexercise}{4}{25}
|
||||||
|
What is $(0.125)^{-2/3}$?
|
||||||
|
|
||||||
|
\begin{equation}
|
||||||
|
b^{p/q} = \sqrt[q]{b^p}
|
||||||
|
\end{equation}
|
||||||
|
|
||||||
|
\begin{myanswer}
|
||||||
|
\begin{align*}
|
||||||
|
0.125^{-2/3} &= \left(\frac{1}{8}\right)^{-2/3} \\
|
||||||
|
&= 8^{2/3} \\
|
||||||
|
&= \sqrt[3]{8^2} \\
|
||||||
|
&= \sqrt[3]{64} \\
|
||||||
|
&= 4
|
||||||
|
\end{align*}
|
||||||
|
\end{myanswer}
|
||||||
|
\end{myexercise}
|
||||||
|
|
||||||
|
\begin{myexercise}{5}{25}
|
||||||
|
|
||||||
|
Defining real numbers in terms of binary expansion.
|
||||||
|
|
||||||
|
\begin{myanswer}
|
||||||
|
\[
|
||||||
|
x = \sum_{i=0}^{\infty} a_i \cdot 2^{-i}
|
||||||
|
\]
|
||||||
|
\end{myanswer}
|
||||||
|
\end{myexercise}
|
||||||
|
|
||||||
|
\begin{myexercise}{6}{25}
|
||||||
|
Let $x = m + 0.d_1d_2\dots$ and $y = n + 0.e_1e_2\dots$.
|
||||||
|
Give a rule for determining whether $x = y$, $x < y$, $x > y$ based on the decimal representation.
|
||||||
|
|
||||||
|
\begin{myanswer}
|
||||||
|
\begin{algorithm}[H]
|
||||||
|
\begin{algorithmic}[1]
|
||||||
|
\Procedure {LowerGreaterEqual} {$X$, $Y$}
|
||||||
|
\Comment{$X$ and $Y$ are arrays of decimal of numbers $x$ and $y$ respectively.}
|
||||||
|
\State $i \gets 0$
|
||||||
|
\While {$i < X.length$ and $i < Y.length$}
|
||||||
|
\If{$X[i] == Y[i]$}
|
||||||
|
\State $i \gets i+1$
|
||||||
|
\ElsIf {$X[i] > Y[i]$}
|
||||||
|
\State \Return \texttt{GREATER}
|
||||||
|
\Else
|
||||||
|
\State \Return \texttt{LOWER}
|
||||||
|
\EndIf
|
||||||
|
\EndWhile
|
||||||
|
\If{$X.length > Y.length$}
|
||||||
|
\State \Return \texttt{GREATER}
|
||||||
|
\ElsIf{$X.length < Y.length$}
|
||||||
|
\State \Return \texttt{LOWER}
|
||||||
|
\Else
|
||||||
|
\State \Return \texttt{EQUALS}
|
||||||
|
\EndIf
|
||||||
|
\EndProcedure
|
||||||
|
\end{algorithmic}
|
||||||
|
\caption{Greater, Lower or Equal Decimal representation}
|
||||||
|
\end{algorithm}
|
||||||
|
\end{myanswer}
|
||||||
|
\end{myexercise}
|
||||||
|
|
||||||
|
\begin{myexercise}{7}{25}
|
||||||
|
Given that $x$ and $y$ are integers. \textit{Prove} the laws of exponents, from definition given by Eq. \ref{eq:rule-of-exponents-2} from Eq. \ref{eq:rule-of-exponents}.
|
||||||
|
|
||||||
|
\begin{eqnarray}
|
||||||
|
b^0 = 1, & b^n = b^{n-1} b \quad \text{ if $n > 0$}, & b^n = b^{n+1} / b \quad \text{ if $n < 0$}.
|
||||||
|
\label{eq:rule-of-exponents}
|
||||||
|
\end{eqnarray}
|
||||||
|
|
||||||
|
\begin{eqnarray}
|
||||||
|
b^{x+y} = b^x b^y, & (b^x)^y = b^{xy},
|
||||||
|
\label{eq:rule-of-exponents-2}
|
||||||
|
\end{eqnarray}
|
||||||
|
|
||||||
|
\begin{myanswer}
|
||||||
|
We will prove this in $\Reals$
|
||||||
|
Let us recall the following formula:
|
||||||
|
\begin{equation}
|
||||||
|
b^x = e^{x \ln b}
|
||||||
|
\end{equation}
|
||||||
|
and
|
||||||
|
\begin{equation}
|
||||||
|
\exp(x + y) = \exp(x) \exp(y)
|
||||||
|
\end{equation}
|
||||||
|
|
||||||
|
Thus it comes:
|
||||||
|
\begin{align*}
|
||||||
|
b^{x + y} &= e^{(x + y) \ln b} \\
|
||||||
|
&= e^{x \ln b + y \ln b} \\
|
||||||
|
&= e^{x \ln b} e^{y \ln b} \\
|
||||||
|
&= b^x b^y
|
||||||
|
\end{align*}
|
||||||
|
|
||||||
|
Similarly for $(a^x)^y$:
|
||||||
|
\begin{align*}
|
||||||
|
(a^x)^y &= e^{y \ln (a^x)} \\
|
||||||
|
&= e^{xy \ln a} \\
|
||||||
|
&= a^{xy}
|
||||||
|
\end{align*}
|
||||||
|
\end{myanswer}
|
||||||
|
\end{myexercise}
|
||||||
|
|
||||||
|
|
|
||||||
Binary file not shown.
|
|
@ -0,0 +1,105 @@
|
||||||
|
<?xml version="1.0" encoding="UTF-8"?>
|
||||||
|
<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" width="612" height="792" viewBox="0 0 612 792">
|
||||||
|
<defs>
|
||||||
|
<g>
|
||||||
|
<g id="glyph-0-0">
|
||||||
|
</g>
|
||||||
|
<g id="glyph-0-1">
|
||||||
|
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||||||
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||||||
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||||||
|
</g>
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||||||
|
<g id="glyph-0-3">
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||||||
|
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|
||||||
|
</g>
|
||||||
|
<g id="glyph-0-4">
|
||||||
|
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||||||
|
</g>
|
||||||
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<g id="glyph-0-5">
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||||||
|
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||||||
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||||||
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||||||
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||||||
|
</g>
|
||||||
|
<g id="glyph-0-7">
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\documentclass{article}
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|
\usepackage{graphicx}
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|
\usepackage{caption}
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|
\begin{document}
|
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|
\begin{figure}
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||||||
|
\centering
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||||||
|
\includegraphics[scale=1]{sum_odd_numbers_to_square}
|
||||||
|
\setcounter{figure}{2}
|
||||||
|
\caption{The sum of odd numbers is a square}
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|
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\end{document}
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% Draw a 6 by 6 grid of squares, with thicker line on outer borders.
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\documentclass[margin=0.5cm]{standalone}
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\usepackage{tikz}
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\usetikzlibrary{calc}
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% Draw grid
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% Draw thicker lines on outer borders
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% Draw for each square with bottom left corner at (0, 0) to (6, 6) the thicker border
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\draw[very thick] (0, 0) -- (\i, 0) -- (\i, \i) -- (0, \i) -- (0, 0);
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% For each $i$-th odd number from 1 to 11, add a node at the top right corner of the $i$-th square
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||||||
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\end{document}
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||||||
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@ -1 +1,7 @@
|
||||||
\chapter*{Introduction}
|
\chapter*{Introduction}
|
||||||
|
|
||||||
|
|
||||||
|
This document gathers some random solution to the problem from the \textit{Art of Computer Programming} by Donald E. KNUTH.
|
||||||
|
It should not be considered as a reference, but as a personal attempt to solve some of the exercises, there for the answer should not consider the solution as correct unless he find it so.
|
||||||
|
|
||||||
|
The answer are presented with page numbers from the third edition of the AOCP.
|
||||||
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||||||
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@ -0,0 +1,3 @@
|
||||||
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\DeclareMathOperator{\Reals}{\mathbb{R}}
|
||||||
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\DeclareMathOperator{\Integers}{\mathbb{Z}}
|
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\DeclareMathOperator{\Naturals}{\mathbb{N}}
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||||||
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@ -1,5 +1,5 @@
|
||||||
% makeindex style file created by the glossaries package
|
% makeindex style file created by the glossaries package
|
||||||
% for document 'main' on 2023-1-3
|
% for document 'main' on 2023-2-27
|
||||||
actual '?'
|
actual '?'
|
||||||
encap '|'
|
encap '|'
|
||||||
level '!'
|
level '!'
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||||||
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|
||||||
BIN
tex/main.pdf
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@ -1,44 +1,23 @@
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||||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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|
||||||
% The LuaLaTeX Chameleon Book Template
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||||||
% LaTeX Template
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||||||
% Version 0.0.0 (2022-12-05)
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|
||||||
%
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||||||
% Authors:
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|
||||||
% Samuel ORTION <samuel+dev@ortion.fr>
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||||||
% Compiling this template:
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||||||
% This template uses biber for its bibliography and makeindex for its index.
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% When you first open the template, compile it from the command line with the
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|
||||||
% commands below to make sure your LaTeX distribution is configured correctly:
|
|
||||||
%
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||||||
% 1) lualatex main
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|
||||||
% 2) makeindex main.idx -s indexstyle.ist
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% 3) biber main
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% 3) makeglossaries main
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% 4) lualatex main
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||||||
% 5) lualatex main
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||||||
%
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|
||||||
% License:
|
|
||||||
% MIT
|
|
||||||
%
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|
||||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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||||||
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||||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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||||||
% PREAMBLE
|
% PREAMBLE
|
||||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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||||||
\documentclass[
|
\documentclass[
|
||||||
11pt,
|
11pt,
|
||||||
fleqn,
|
fleqn,
|
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a4paper
|
a4paper,
|
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oneside
|
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]{TheLuaChameleonBook}
|
]{TheLuaChameleonBook}
|
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|
||||||
% Book PDF metadata
|
% Book PDF metadata
|
||||||
\hypersetup{
|
\hypersetup{
|
||||||
pdftitle={The Art of Computer Programming Vol. 1 - Some Exercises},
|
pdftitle={The Art of Computer Programming Vol. 1 - Some Exercises},
|
||||||
pdfauthor={Samuel ORTION},
|
pdfauthor={Samuel ORTION},
|
||||||
pdfsubject={An Attempt to solve some of the exercises from the Book of Donald E. KNUTH},
|
pdfsubject={An Attempt to solve some of the exercises from the Book of Donald E. KNUTH},
|
||||||
pdfkeywords={computer science, mathematics, algorithmics},
|
pdfkeywords={computer science, mathematics, algorithmic},
|
||||||
pdfcreator={LaTeX},
|
pdfcreator={LaTeX},
|
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}
|
}
|
||||||
|
|
||||||
\definecolor{primarycolor}{RGB}{0, 128, 0}
|
\definecolor{primarycolor}{RGB}{0, 128, 0}
|
||||||
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|
@ -48,6 +27,7 @@
|
||||||
\makeindex
|
\makeindex
|
||||||
|
|
||||||
\input{preamble.tex}
|
\input{preamble.tex}
|
||||||
|
\input{definitions.tex}
|
||||||
\setcounter{tocdepth}{2}
|
\setcounter{tocdepth}{2}
|
||||||
|
|
||||||
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||||||
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@ -87,7 +67,7 @@
|
||||||
|
|
||||||
\noindent \texttt{\href{https://samuel.ortion.fr/}{samuel.ortion.fr}}\\ % URL
|
\noindent \texttt{\href{https://samuel.ortion.fr/}{samuel.ortion.fr}}\\ % URL
|
||||||
|
|
||||||
\noindent Licensed under the Creative Commons Attribution-NonCommercial 4.0 License.
|
\noindent Answers Licensed under the Creative Commons Attribution-NonCommercial 4.0 License.
|
||||||
% You may not use this file except in compliance with the License. You may obtain a copy of the License at \url{https://creativecommons.org/licenses/by-nc-sa/4.0}. Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an \textsc{``as is'' basis, without warranties or conditions of any kind}, either express or implied. See the License for the specific language governing permissions and limitations under the License.\\ % License information, replace this with your own license (if any)
|
% You may not use this file except in compliance with the License. You may obtain a copy of the License at \url{https://creativecommons.org/licenses/by-nc-sa/4.0}. Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an \textsc{``as is'' basis, without warranties or conditions of any kind}, either express or implied. See the License for the specific language governing permissions and limitations under the License.\\ % License information, replace this with your own license (if any)
|
||||||
|
|
||||||
% \noindent \textit{First printing, October 2022} % Printing/edition date
|
% \noindent \textit{First printing, October 2022} % Printing/edition date
|
||||||
|
|
|
||||||
|
|
@ -2,65 +2,61 @@
|
||||||
\usepackage{xstring}
|
\usepackage{xstring}
|
||||||
|
|
||||||
\newcommand{\unfinished}{
|
\newcommand{\unfinished}{
|
||||||
\textbf{\textcolor{orange!90}{to be continued...}}
|
\textbf{\textcolor{orange!90}{to be continued...}}
|
||||||
}
|
}
|
||||||
|
|
||||||
\newcommand{\unresolved}{
|
\newcommand{\unresolved}{
|
||||||
\textbf{\textcolor{red!90}{unresolved}}
|
\textbf{\textcolor{red!90}{unresolved}}
|
||||||
}
|
}
|
||||||
|
|
||||||
% Create a new environment {exercise} to refer to exercises from the KNUTH book.
|
% Create a new environment {exercise} to refer to exercises from the KNUTH book.
|
||||||
% Add a TODO switch option to show unresolved exercises
|
|
||||||
\newenvironment{myexercise}[2]
|
\newenvironment{myexercise}[2]
|
||||||
{
|
{
|
||||||
\addcontentsline{toc}{subsubsection}{Exercise #1}
|
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|
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\begin{minipage}{\textwidth}
|
\begin{flushleft}
|
||||||
\begin{flushleft}
|
{
|
||||||
{
|
\Large
|
||||||
\Huge
|
\textbf{Exercise #1.} (p. #2)
|
||||||
\textbf{Exercise #1.} (p. #2)
|
}
|
||||||
}
|
\end{flushleft}
|
||||||
\end{flushleft}
|
|
||||||
}
|
}
|
||||||
{
|
{
|
||||||
\end{minipage}
|
|
||||||
}
|
}
|
||||||
|
|
||||||
% Create a new environment for big quote with `primarycolor' coloured quote sign
|
% Create a new environment for big quote with `primarycolor' coloured quote sign
|
||||||
\newenvironment{myquote}
|
\newenvironment{myquote}
|
||||||
{
|
{
|
||||||
\begin{minipage}{\textwidth}
|
\begin{flushleft}
|
||||||
\begin{flushleft}
|
{
|
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{
|
|
||||||
|
|
||||||
\fontsize{75pt}{3pt}\selectfont
|
\fontsize{75pt}{3pt}\selectfont
|
||||||
\textcolor{primarycolor}{\textquotedblleft}
|
\textcolor{primarycolor}{\textquotedblleft}
|
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}
|
}
|
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}
|
}
|
||||||
{
|
{
|
||||||
\begin{flushright}
|
\begin{flushright}
|
||||||
{
|
{
|
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\textcolor{primarycolor}{\textquotedblright}
|
\textcolor{primarycolor}{\textquotedblright}
|
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|
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|
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|
|
||||||
\end{flushright}
|
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|
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\end{flushleft}
|
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|
||||||
\end{minipage}
|
|
||||||
}
|
}
|
||||||
|
|
||||||
\newenvironment{myanswer}
|
\newenvironment{myanswer}
|
||||||
{
|
{
|
||||||
\begin{minipage}{\textwidth}
|
\begin{itshape}
|
||||||
\begin{itshape}
|
\begin{flushleft}
|
||||||
\begin{flushleft}
|
{
|
||||||
{
|
\textbf{My Answer}:
|
||||||
\Large
|
}
|
||||||
\textbf{My Answer}:
|
|
||||||
}
|
|
||||||
}
|
}
|
||||||
{
|
{
|
||||||
\end{flushleft}
|
\end{flushleft}
|
||||||
\end{itshape}
|
\end{itshape}
|
||||||
\end{minipage}
|
|
||||||
}
|
}
|
||||||
|
|
||||||
|
\usepackage{algorithmicx}
|
||||||
|
\usepackage{algorithm}
|
||||||
|
\usepackage{algpseudocodex}
|
||||||
|
|
||||||
|
|
|
||||||
Loading…
Reference in New Issue